If `f,g,` and `h` are differentiable functions of `x` and `(delta)`=`|[f,g,h], [(xf)',(xg)',(xh)'],[(x^2f)' ',(x^2g)' ',(x^2h)' ']|` prove that `delta^(prime)`=`|[f,g,h],[f',g', h],[ (x^3f' ')',(x^3g' ')',(x^3h ' ')']|`

Explanation

` (xf) =xf' +f " and " (x^(2)f)''`
`=[2xf+ x^(2) f]'`
` =2f +4xf'+ x^(2) f''`
`rArr Delta =| {:( f,,g,,h),(xf+f,,xg'+g,,xh'+h),(2f+4xf+x^(2)f'',,2g+4xg'+x^(2)g'',,2h+4xh+x^(2)h''):}| `
`R_(2) to R_(2) -R_(1) " and then " R_(3) to R_(3) -4R_(2) -2R_(1)`
`rArr Delta = |{:(f,,g,,h),(xf,,xg,,xh),(x^(2)f,,x^(2)g,,x^(2)h):}|`
`Delta = |{:(f,,g,,h),(xf,,xg,,xh),(x^(3)f,,x^(3)g,,x^(3)h):}|`
`rArr (dDelta)/(dx) = |{:(f',,g',,h'),(f',,g',,h'),(x^(3)f'',,x^(3)g'',,x^(3) h''):}|+|{:(f,,g,,h),(f'',,g'',,h''),(x^(3)f'',,x^(3)g'',,x^(3)h''):}|`
`+|{:(f,,g,,h),(f',,g',,h'),((x^(3)f'')',,(x^(3)g'')',,(x^(3)h'')'):}|`
`=0+0 + |{:(f,,g,,h),(f',,g',,h'),((x^(3)f'')',,(x^(3)g'')',,(x^(3)h'')'):}|`